3.478 \(\int e^{4 \tanh ^{-1}(a x)} (c-\frac{c}{a x})^2 \, dx\)

Optimal. Leaf size=27 \[ -\frac{c^2}{a^2 x}+\frac{2 c^2 \log (x)}{a}+c^2 x \]

[Out]

-(c^2/(a^2*x)) + c^2*x + (2*c^2*Log[x])/a

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Rubi [A]  time = 0.0961591, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6131, 6129, 43} \[ -\frac{c^2}{a^2 x}+\frac{2 c^2 \log (x)}{a}+c^2 x \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])*(c - c/(a*x))^2,x]

[Out]

-(c^2/(a^2*x)) + c^2*x + (2*c^2*Log[x])/a

Rule 6131

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 + (c*x)/d)
^p*E^(n*ArcTanh[a*x]))/x^p, x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]

Rule 6129

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[(u*(1 + (d*x)/c)
^p*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ
[p] || GtQ[c, 0])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{4 \tanh ^{-1}(a x)} \left (c-\frac{c}{a x}\right )^2 \, dx &=\frac{c^2 \int \frac{e^{4 \tanh ^{-1}(a x)} (1-a x)^2}{x^2} \, dx}{a^2}\\ &=\frac{c^2 \int \frac{(1+a x)^2}{x^2} \, dx}{a^2}\\ &=\frac{c^2 \int \left (a^2+\frac{1}{x^2}+\frac{2 a}{x}\right ) \, dx}{a^2}\\ &=-\frac{c^2}{a^2 x}+c^2 x+\frac{2 c^2 \log (x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.0936694, size = 29, normalized size = 1.07 \[ -\frac{c^2}{a^2 x}+\frac{2 c^2 \log (a x)}{a}+c^2 x \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(4*ArcTanh[a*x])*(c - c/(a*x))^2,x]

[Out]

-(c^2/(a^2*x)) + c^2*x + (2*c^2*Log[a*x])/a

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Maple [A]  time = 0.034, size = 28, normalized size = 1. \begin{align*} -{\frac{{c}^{2}}{{a}^{2}x}}+x{c}^{2}+2\,{\frac{{c}^{2}\ln \left ( x \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^2,x)

[Out]

-c^2/a^2/x+x*c^2+2*c^2*ln(x)/a

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Maxima [A]  time = 0.95113, size = 36, normalized size = 1.33 \begin{align*} c^{2} x + \frac{2 \, c^{2} \log \left (x\right )}{a} - \frac{c^{2}}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^2,x, algorithm="maxima")

[Out]

c^2*x + 2*c^2*log(x)/a - c^2/(a^2*x)

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Fricas [A]  time = 2.02426, size = 65, normalized size = 2.41 \begin{align*} \frac{a^{2} c^{2} x^{2} + 2 \, a c^{2} x \log \left (x\right ) - c^{2}}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^2,x, algorithm="fricas")

[Out]

(a^2*c^2*x^2 + 2*a*c^2*x*log(x) - c^2)/(a^2*x)

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Sympy [A]  time = 0.328515, size = 26, normalized size = 0.96 \begin{align*} \frac{a^{2} c^{2} x + 2 a c^{2} \log{\left (x \right )} - \frac{c^{2}}{x}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+1)**2*(c-c/a/x)**2,x)

[Out]

(a**2*c**2*x + 2*a*c**2*log(x) - c**2/x)/a**2

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Giac [A]  time = 1.17434, size = 38, normalized size = 1.41 \begin{align*} c^{2} x + \frac{2 \, c^{2} \log \left ({\left | x \right |}\right )}{a} - \frac{c^{2}}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^2,x, algorithm="giac")

[Out]

c^2*x + 2*c^2*log(abs(x))/a - c^2/(a^2*x)